Two Envelopes Paradox under Continuous Prior

Scenario In quantitative trading and quantitative research, candidates are frequently evaluated on probability modeling, specifically regarding Conditional Expectation / Integrability.

Your Task
You are given two indistinguishable envelopes. One contains \X$and the other contains$$2X$. You open one envelope and observe$$y$. (1) The classic fallacy asserts: "The other envelope is equally likely to contain$y/2$or$2y$, so its expected value is$\frac{1}{2}(y/2) + \frac{1}{2}(2y) = 1.25y > y$, so you should always switch." Prove mathematically why this argument is fallacious. (2) Formulate the condition on the prior density$f_X(x)$under which switching yields a strictly positive expected gain for all observed$y$. Prove that no proper probability distribution with finite mean can satisfy this condition everywhere.

Constraints

• All variables are defined on a standard probability space $(\Omega, \mathcal{F}, \mathbb{P})$.
• Analytical expressions must be in closed form, and empirical simulations should converge within standard statistical tolerances.